Machine-translated from Chinese.
YMSC公开课
定义
给定有限集$V,$ 基本$p$-道路为含有$V$中$p+1$个点的序列
$$
e_{i_0\cdots i_p}=\{i_0,\cdots,i_p\}.
$$
固定数域$\mathbb{K},$
$\Lambda_p$为所有基本$p$道路的$\mathbb{K}$-线性组合,
其中的元素称为$p$-道路.
定义$\partial:\Lambda_p\rightarrow \Lambda_{p-1}:$
$$
\partial e_{i_0\cdots i_p}=\sum_{q=0}^p(-1)^qe_{i_0\cdots \widehat i_q\cdots i_p}, \quad p\ge 1,
$$
$p=0$时定义$\Lambda_{-1}=\{0\},$ $\partial=0.$
这样$(\Lambda_\ast ,\partial)$构成链复形. 由基本计算, 我们有如下引理.
引理 1. $\partial^2=0.$
我们称基本$p$-道路$i_0\cdots i_p$是正则的, 若$i_k\neq i_{k+1},$
$\,\forall\,k=0,\cdots,p-1.$ 反之成为非正则的.
记$I_p$为所有非正则基本$p$-道路张成的$\Lambda_p$中的子空间.
我们有$\partial I_p\subset I_{p-1}.$ 由此可合理定义$R_p=\Lambda_p/I_p,$
使得有诱导运算$\partial:R_p\rightarrow R_{p-1}.$
称$R_p$中元素为正则化的道路.
若$V$为有向图, 进一步称正则基本$p$-道路$i_0\cdots i_p$是准许的,
若$i_k\rightarrow i_{k+1},$ $\,\forall\,k=0,\cdots,p-1.$
记$A_p$为全体准许正则基本$p$-道路生成的子空间. 一般的,
$\partial A_p\not\subset A_{p-1}.$
定义$A_p$的子空间$\Omega_p:$
$$
\Omega_p:=\{v\in A_p\mid \partial v\in A_{p-1}\}.
$$
此时有$\partial \Omega_p\subset \Omega_{p-1}$.
$\Omega_p$中的元素称为$\partial$-不变$p$-道路.
$A_0=\left<{}e_i\right>,$
$A_1=\left<{}e_{ij}\mid i\rightarrow j\right>,$
$A_2=\left<{}e_{ijk}\mid i\rightarrow j\rightarrow k\right>.$
那么$\Omega_0=A_0,$ $\Omega_1=A_1,$ 而一般的$\Omega_2\subset A_2.$
一些例子是若$a\rightarrow b\rightarrow c$且$a\rightarrow c,$
那么三角形$e_{abc}\in \Omega_2;$
若$a\rightarrow b\rightarrow c$且$a\rightarrow b'\rightarrow c,$
但$a\not\rightarrow c,$ 那么也有矩形$e_{abc}-e_{ab'c}\in \Omega_2;$
还有双箭头$e_{aba}\in \Omega_2.$
命题 2. $|\Omega_p|\le 1$ $\Rightarrow$ $|\Omega_n|=0,$ $\,\forall\,n> p.$
称序列$0,1,\cdots,p$为$p$-单形, 若其满足$i<j$ $\Leftrightarrow$
$i\rightarrow j.$ $p$-单形是落在$\Omega_p$中的.
道路同调
定义$H_p=\ker \partial_{p}/\operatorname{Im}\partial_{p+1},$
$\partial_p:=\partial|_{\Omega_p}.$ Betti数 $\beta_p:=|H_p|.$
若$\Omega_\ast $是有限的, 即$\Omega_p=\{0\}$对$p$充分大, 那么示性数
$$
\chi:=\sum_{p=0}^\infty (-1)^p |\Omega_p|=\sum_{p=0}^\infty (-1)^p \beta_p.
$$
命题 3. 若$X,Y$为两个无交的有向图, 那么$\beta_p(X\sqcup Y)=\beta_p(X)+\beta_p(Y).$
这一命题来自于$\Omega_p(X\sqcup Y)=\Omega_p(X)\oplus \Omega_p(Y).$
进一步, 有
命题 4. $\beta_0(G)$为$G$的连通分支个数.
只需证明若$G$是连通的, $\beta_0(G)=1.$
这里连通指任意两点$i,j$之间存在$i$到$j$的道路. 由此即得.
称$a,b$构成半有向边, 若$a\not\rightarrow b,$ 但存在$c,$
使得$a\rightarrow c\rightarrow b.$
命题 5. $|\Omega_2|=|A_2|-s,$ $s$为$G$中半有向边个数. 事实上, $\Omega_2$由所有的三角形, 矩形, 双箭头张成.
文章最后更新于 2024-02-26 15:27:06
YMSC open courses
definition
Given a finite set $V,$, the basic $p$-road is a sequence containing $p+1$ points in $V$
$$
e_{i_0\cdots i_p}=\{i_0,\cdots,i_p\}.
$$
Fixed number field$\mathbb{K},$
$\Lambda_p$ is the $\mathbb{K}$-linear combination of all basic $p$ roads,
The elements in it are called $p$-roads.
Definition$\partial:\Lambda_p\rightarrow \Lambda_{p-1}:$
$$
\partial e_{i_0\cdots i_p}=\sum_{q=0}^p(-1)^qe_{i_0\cdots \widehat i_q\cdots i_p}, \quad p\ge 1,
$$
Define $\Lambda_{-1}=\{0\},$ $\partial=0.$ when $p=0$
In this way $(\Lambda_\ast ,\partial)$ forms a chain complex. From basic calculations, we have the following lemma.
Lemma 1. $\partial^2=0.$
We say that the basic $p$-road $i_0\cdots i_p$ is regular if $i_k\neq i_{k+1},$
$\,\forall\,k=0,\cdots,p-1.$ Otherwise it becomes irregular.
Let $I_p$ be the subspace in $\Lambda_p$ spanned by all non-regular basic $p$-roads.
We have $\partial I_p\subset I_{p-1}.$ from which we can reasonably define $R_p=\Lambda_p/I_p,$
Enables induced operations$\partial:R_p\rightarrow R_{p-1}.$
The elements in $R_p$ are called regularized roads.
If $V$ is a directed graph, it is further said that the regular basic $p$-road $i_0\cdots i_p$ is allowed,
If $i_k\rightarrow i_{k+1},$ $\,\forall\,k=0,\cdots,p-1.$
Let $A_p$ be the subspace generated by the universally allowed regular basis $p$-road. Generally,
$\partial A_p\not\subset A_{p-1}.$
Define the subspace $\Omega_p:$ of $A_p$
$$
\Omega_p:=\{v\in A_p\mid \partial v\in A_{p-1}\}.
$$
At this time there is $\partial \Omega_p\subset \Omega_{p-1}$.
The elements in $\Omega_p$ are called $\partial$-invariant $p$-roads.
$A_0=\left<{}e_i\right>,$
$A_1=\left<{}e_{ij}\mid i\rightarrow j\right>,$
$A_2=\left<{}e_{ijk}\mid i\rightarrow j\rightarrow k\right>.$
Then $\Omega_0=A_0,$ $\Omega_1=A_1,$ and normal $\Omega_2\subset A_2.$
Some examples are if $a\rightarrow b\rightarrow c$ and $a\rightarrow c,$
Then the triangle $e_{abc}\in \Omega_2;$
If $a\rightarrow b\rightarrow c$ and $a\rightarrow b'\rightarrow c,$
But $a\not\rightarrow c,$ Then there are also rectangles $e_{abc}-e_{ab'c}\in \Omega_2;$
There are also double arrows $e_{aba}\in \Omega_2.$
Proposition 2. $|\Omega_p|\le 1$ $\Rightarrow$ $|\Omega_n|=0,$ $\,\forall\,n> p.$
The sequence $0,1,\cdots,p$ is called $p$-simplex if it satisfies $i<j$ $\Leftrightarrow$
$i\rightarrow j.$ $p$ - The simplex falls within $\Omega_p$.
road synchrony
Definition$H_p=\ker \partial_{p}/\operatorname{Im}\partial_{p+1},$
$\partial_p:=\partial|_{\Omega_p}.$ Betti number $\beta_p:=|H_p|.$
If $\Omega_\ast $ is finite, that is, $\Omega_p=\{0\}$ is sufficiently large to $p$, then the indicator number
$$
\chi:=\sum_{p=0}^\infty (-1)^p |\Omega_p|=\sum_{p=0}^\infty (-1)^p \beta_p.
$$
Proposition 3. If $X,Y$ is two non-intersecting directed graphs, then $\beta_p(X\sqcup Y)=\beta_p(X)+\beta_p(Y).$
This proposition comes from $\Omega_p(X\sqcup Y)=\Omega_p(X)\oplus \Omega_p(Y).$
further, there are
Proposition 4. $\beta_0(G)$ is the number of connected components of $G$.
Just prove that if $G$ is connected, $\beta_0(G)=1.$
Connectivity here means that there is a road from $i$ to $j$ between any two points $i,j$. This is obtained.
$a,b$ is said to form a semi-directed edge, if $a\not\rightarrow b,$ exists but $c,$ exists
Make$a\rightarrow c\rightarrow b.$
Proposition 5. $|\Omega_2|=|A_2|-s,$ $s$ is the number of semi-directed edges in $G$. In fact, $\Omega_2$ is composed of all triangles, rectangles, and double arrows.
The article was last updated on 2024-02-26 15:27:06
